Question: Is there a polynomial map from ℝn to ℝn under which the image of the positive orthant (the set of points with all coordinates positive) is all of ℝn?

Some observations:

My intuition is that the answer must be 'no'... but I confess my intuition for this sort of geometric problem is not very well-developed.

Of course it is relatively easy to show that the answer is 'no' when n=1. (In fact it seems like a nice homework problem for some calculus students.) But I can't seem to get any traction for n>1.

This feels like the sort of thing that should have an easy proof, but then I remember feeling that way the first time I saw the Jacobian conjecture... now I'm wary of statements about polynomial maps of ℝn!

1 Answer
1

The map $z\in\mathbb C\mapsto z^4\in\mathbb C$, when written out in coordinates, is a polynomial map which sends the closed first quadrant to the whole of $\mathbb R^2$---and by considering cartesian products you get the same for $\mathbb R^{2n}=\mathbb C^n$.

Later: as observed in a comment by Charles, this can be turned into a solution for the open quadrant by composing with a translation, as in $z\in\mathbb C\mapsto (z-z_0)^4\in\mathbb C$ with $z_0$ in the open first quadrant.

This yields maps for all even dimensions. Is there a simple answer to the odd-dimensional case?
–
j.c.Jan 15 '10 at 19:44

7

Yes, because you can use this trick for any pair of coordinates in $\mathbb{R}^n$, regardless of the parity of $n$, and then compose until positivity has been eliminated for all coordinates.
–
Greg KuperbergJan 15 '10 at 19:45